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neverforgetvivistee

  • 4 years ago

solving rational equations: x+2 x - 1 4x+2 ---- = ---- - ----- x x x^2-3x I know how to make them all have a common denominator x(x-3) but when it comes to solving I get x^2+5x+6=x^2-4x-3x-3x-2 which then comes up to be x^2+3x-11 This isn't right. What did I do wrong?

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  1. anonymous
    • 4 years ago
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    maybe it would be easier to multiply both sides by \[x(x-3)\]

  2. neverforgetvivistee
    • 4 years ago
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    i did

  3. neverforgetvivistee
    • 4 years ago
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    but i can't figure out what i did wrong because i get the wrong answer... when i multiply x+2 and x-3 i get x^2+5x+5 when i multiply x-1 and x-3 i get x^2-4x-3

  4. anonymous
    • 4 years ago
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    you get \[(x+2)(x-3)=(x-1)(x-3)-4x+2\] i htink

  5. anonymous
    • 4 years ago
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    hold on \[(x+2)(x-3)=x^2-x-6\] ,maybe that is the mistake

  6. anonymous
    • 4 years ago
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    also \[(x-1)(x-3)=x^2-4x+3\]

  7. neverforgetvivistee
    • 4 years ago
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    oh!!!!!! i put (x+2) and postive (x+3)!!

  8. neverforgetvivistee
    • 4 years ago
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    that was a waste of time i was looking just at how i solved it and not how i multiplied it. Thanks

  9. anonymous
    • 4 years ago
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    \[(x+2)(x-3)=(x-1)(x-3)-4x+2\] \[x^2-x-6=x^2-4x+3-4x+2\] \[x^2-x-6=x^2-8x+5\] etc

  10. anonymous
    • 4 years ago
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    yw

  11. neverforgetvivistee
    • 4 years ago
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    when you're subtracting rationals, do you make it into a plus sign and make EVERYTHING in the numerator of the second negative, or just make 1 number negative? for examples x-1 4x+2 ---- - ----- x(x-3) x(x-3) would you make it into a plus sign, and then make 4x and 2 negative? -4x-2

  12. neverforgetvivistee
    • 4 years ago
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    bump

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